Answer:
68.585m/sec , 779.1 N
Explanation:
To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.
From centripetal motion.
F =( mv^2)/2
But since we are dealing with weightlessness
r = 480m
g = 9.8m/s^2
M also cancels, so forget M.
V^2 = Fr
V = √ Fr
V =√ (9.8 x 480) = 4704
= 68.585m/sec.
b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960
= 4.9m/sec^2.
Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)
159kg × ( 9.8-4.9)
159kg × 4.9
= 779.1N
A) The speed of the scooter at which the driver will feel weightlessness is;
v = 68.586 m/s
B) The apparent weight of both the driver and the scooter at the top of the hill is;
F_net = 779.1 N
We are given;
Mass; m = 159 kg
Radius; r = 480 m
A) Since it's motion about a circular hill, it means we are dealing with centripetal force.
Formula for centripetal force is given as;
F = mv²/r
Now, we want to find the speed of the scooter if the driver feels weightlessness.
This means that the centripetal force would be equal to the gravitational force.
Thus;
mg = mv²/r
m will cancel out to give;
v²/r = g
v² = gr
v = √(gr)
v = √(9.8 × 480)
v = √4704
v = 68.586 m/s
B) Now, he is travelling with speed of;
v = 68.586 m/s
And the radius is 2r
Let's first find the centripetal acceleration from the formula; α = v²/r
Thus; α = 4704/(2 × 480)
α = 4.9 m/s²
Now, since he has encountered a hill with a radius of 2r up the slope, it means that the apparent weight will now be;
F_app = m(g - α)
F_net = 159(9.8 - 4.9)
F_net = 779.1 N
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If a bicycle starts from rest and is pedaled normally until the bike is moving at 6 meters per second across level ground, what kinds of energy have its tires been given? (Select all that apply) g
Answer: Translational Kinetic Energy
Rotational Kinetic Energy
Explanation:
An object has translational kinetic energy when it is undergoing through a linear displacement.
Rotational energy is kinetic energy due to the rotation of an object .
Here the wheel of bicycle undergoes both translational and rotational kinetic energy has it moves with linear displacement with rotation in it.
Hence, the tires have been two kinds of energy : translational and rotational kinetic energy
A moon orbits a planet along an elliptical path. Which describes the location of the planet within the ellipse
Must be at a focus.
----------------
Hope this helps!
Brainliest would be great!
----------------
With all care,
07x12
One of the two foci of the elliptical route that the moon follows is the planet. The planet will be positioned at one of the two focal points (foci) of an elliptical orbit. The other focus is still vacant.
The two foci are connected by the ellipse's major axis, and for any given ellipse, their separation is constant. The mass of the planet and the mass of the object it orbits—for example, a star if it's a planet in a solar system—are used to calculate the planet's position. The moon revolves the planet in an elliptical pattern due to the gravitational forces between the planet and the object it orbits, with the planet itself at one of the foci.
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a car brakes and stops at 10 [m]. While stopping, the friction force of the wheels on the pavement is 400 [N]. Calculate the work done.
Explanation:
Work = force × distance
W = (400 N) (10 m)
W = 4000 J
You indicate that a symbol
is a vector by drawing
A. through the symbol.
B. over the symbol.
c. under the symbol.
D. before the symbol.
Answer:
B. over the symbol.
Explanation:
vectors are represented with a symbol carrying an arrow head with also indicates direction
Will give brainliest ASAP! Please help (1/10 questions, will mark 5 stars and brainliest for all answers if correct)
Answer:
Option (A)
Explanation:
A 20 kg boy chases the butterfly with a speed of 2 meter per second.
Angle at which he runs is 70° North of West.
Therefore, Horizontal component (Vx) directing towards West will be,
Vx = v(Cos70°)
Vy = v(Sin70°)
Since momentum of a body is defined by,
Momentum = Mass × Velocity
Therefore, Westerly component of the momentum will be,
Momentum = 20 × (v)(Cos70°)
= 20 × 2Cos70°
= 13.68
≈ 13.7 kg-meter per second
Therefore, Option (A) will be the answer.
You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after pushing it across some distance d, and the block leaves your hand with a velocity of 0.85 m/s. While you are pushing, the work done by friction between the ice and the ground is 3 Nm (3 J). Assuming that the ice block was stationary before you push it, find d.
Answer: d = 33 cm or 0.33 m
Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:
W = F.d.cosθ
F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.
For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:
W = F.d.cos(0)
W = F.d
To determine d:
d = [tex]\frac{W}{F}[/tex]
d = [tex]\frac{3}{9}[/tex]
d = 0.33 m
The distance d the block ice moved is 33 cm.
EXAMPLE 5 Find the radius of gyration about the x-axis of a homogeneous disk D with density rho(x, y) = rho, center the origin, and radius a. SOLUTION The mass of the disk is m = rhoπa2, so from these equations we have 2 = Ix m = 1 4πrhoa4 rhoπa2 = a2 4 .
Answer:
Radius of gyration = a/2.
Explanation:
So, from the question above I can see that the you are already answering the question and you are stuck up or maybe that's how the problem is set from the start. Do not worry, you are covered in any of the ways. So, from the question we have that;
"The mass of the disk is m = ρπa^2, so from these equations we have y^2 = Ix/m."
(NB: I changed the "rho" word to its symbol).
Thus, the radius of gyration with respect to x-axis = (1/4 πρa^4)/ πρa^2 = a^2/4.
Therefore, the Radius of gyration = a/2.
Suppose that a 0.275 m radius, 500 turn coil produces an average emf of 11800 V when rotated one-fourth of a revolution in 4.42 ms, starting from its plane being perpendicular to the magnetic field.
Required:
Find the magnetic field strength needed to induce an average emf of 10,000 V.
Answer:
The magnetic field strength : 0.372 T
Explanation:
The equation of the induced emf is given by the following equation,
( Equation 1 ) emf = - N ( ΔФ / Δt ) - where N = number of turns of the coil, ΔФ = change in the magnetic flux, and Δt = change in time
The equation for the magnetic flux is given by,
( Equation 2 ) Ф = BA( cos( θ ) ) - where B = magnetic field, A = area, and θ = the angle between the normal and the magnetic field
The area of the circular coil is a constant, as well as the magnetic field. Therefore the change in the magnetic flux is due to the angle between the normal and the magnetic field. Therefore you can expect the equation for the change in magnetic flux to be the same as the magnetic flux, but only that there must be a change in θ.
( Equation 3 ) ΔФ = BA( Δcos( θ ) )
Now as the coil rotates one-fourth of a revolution, θ changes from 0 degrees to 90 degrees. The " change in cos θ " should thus be the following,
Δcos( θ ) = cos( 90 ) - cos( 0 )
= 0 - 1 = - 1
Let's substitute that value in the third equation,
( Substitution of Δcos( θ ) previously, into Equation 3 )
ΔФ = BA( - 1 ) = - BA
Remember the first equation? Well if the change in the magnetic flux = - BA, then through further substitution, the emf should = - N( - BA ) / Δt. In other words,
emf = - N( - BA ) / Δt,
emf = NBA / Δt,
B = ( emf )Δt / NA
Now that we have B, the magnetic field strength, isolated, let's solve for the area of the circular coil and substitute all known values into this equation.
Area ( A ) = πr²,
= π( 0.275 )² = 0.2376 m²,
B = ( 10,000 V )( 4.42 [tex]*[/tex] 10⁻³ s ) / ( 500 )( 0.2376 m² ) = ( About ) 0.372 T
The magnetic field strength : 0.372 T
A source containing a mixture of hydrogen and deuterium atoms emits light at two wavelengths whose mean is 540 nm and whose separation is 0.170 nm. Find the minimum number of lines needed in a diffraction grating that can resolve these lines in the first order.
Answer:
N=3176.5rulling
Explanation:
We were told that the source containing a mixture of hydrogen and deuterium atoms emits light with
wavelengths whose mean is 540 nm
Then λ= 540 nm, but we need to convert to metre which = (540× 10⁻⁹m)
Also whose separation is 0.170 nm, which mean the difference between the wavelength is 0.170 nm then
Δ λ = 0.170 nm the we convert to metre we have. Δλ= 0.170 nm= (0.170×10⁻⁹m)
the formular below can be used to can be used to calculate our minimum number of lines
N= λ /(m Δλ)
Where N is number of fillings i.e number of lines
λ= wavelength
Δλ= difference in wavelength
m=1
Then if we substitute the values we have
,N= (540× 10⁻⁹ m)/[(1)*(0.170× 10⁻⁹m)]
N =3176.5rulling
Therefore, minimum number of lines = =3176.5rulling
Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness
This is note the complete question, the complete question is:
One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.
Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?
Answer: 3.04 D
Explanation:
when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.
An image of 0.47m from the eye will be ( 47 - 2 )
i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.
Also, the image distance will become negative
gap between lense and eye = 2cm
Therefore;
image distance d₁ = - 45cm = - 0.45m
object distance d₀ = 21 - 2 = 19cm = 0.19m
P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)
P = 1/f = 5.26315789 - 2.22222222
P = 1/f = 3.04093567 ≈ 3.04 D
Which of the following statement regarding orbits is true?
A) In a orbit, the satellite and the central body are the two Foci of the eclipse.
B) the orbit of the planet has an elliptical shape with the Sun at one Focus.
C) an elliptical orbit there is one focus and the satellite is located there.
D) the sun and a planet are the two Foci of an orbit
The correct option is option (B)
The orbit of the planets is elliptical with the sun at one of the foci.
Orbit of planets:According to Kepler's Laws:
The orbit of a planet around the Sun is an ellipse, with the Sun in one of the focal points of that ellipse. The planet's orbit lies in a plane, called the orbital plane. The point on the orbit closest to the attracting body is the periapsis. The point farthest from the attracting body is called the apoapsis. As the planet moves in its orbit, the line from the Sun to the planet sweeps a constant area of the orbital plane for a given period of time. This means that the planet moves faster near its perihelion than near its aphelion, because at the smaller distance it needs to trace a greater arc to cover the same area. This law is usually stated as equal areas in equal time.For a given orbit, the ratio of the cube of its semi-major axis to the square of its period is constant.Learn more about Kepler's Laws:
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A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through a potential difference of 2V0, what speed would it gain
Answer:
Explanation:
Let the charge on proton be q .
energy gain by proton in a field having potential difference of V₀
= V₀ q
Due to gain of energy , its kinetic energy becomes 1/2 m v₀²
where m is mass and v₀ is velocity of proton
V₀ q = 1/2 m v₀²
In the second case , gain of energy in electrical field
= 2 V₀q , if v be the velocity gained in the second case
2 V₀q = 1/2 m v²
1/2 m v² = 2 V₀q = 2 x 1/2 m v₀²
mv² = 2 m v₀²
v = √2 v₀
In the 1980s, the term picowave was used to describe food irradiation in order to overcome public resistance by playing on the well-known safety of microwave radiation. Find the energy in MeV of a photon having a wavelength of a picometer.
Answer:
E = 1.24MeV
Explanation:
The photon travels at the speed of light, 3.0 × [tex]10^{8}[/tex] m/s, and given that its frequency = 1 picometer = 1.0 × [tex]10^{-12}[/tex] m.
Its energy can be determined by;
E = hf
= (hc) ÷ λ
where E is the energy, h is the Planck's constant, 6.626 × [tex]10^{-34}[/tex] Js, c is the speed of the light and f is its frequency.
Therefore,
E = (6.626 × [tex]10^{-34}[/tex]× 3.0 × [tex]10^{8}[/tex]) ÷ 1.0 × [tex]10^{-12}[/tex]
= 1.9878 × [tex]10^{-25}[/tex] ÷ 1.0 × [tex]10^{-12}[/tex]
E = 1.9878 × [tex]10^{-13}[/tex] J
But, 1 eV = 1.6 × [tex]10^{-19}[/tex] J. So that;
E = [tex]\frac{1.9878*10^{-13} }{1.6*10^{-19} }[/tex]
= 1242375 eV
∴ E = 1.24MeV
The energy of the photon is 1.24MeV.
A 2.0 kg handbag is released from the top of the Leaning Tower of Pisa, and 55 m before reaching the ground, it carries a speed of 29 m / s. What was the average force of air resistance?
Answer:
4.31 N
Explanation:
Given:
Δy = -55 m
v₀ = 0 m/s
v = -29 m/s
Find: a
v² = v₀² + 2aΔy
(-29 m/s)² = (0 m/s)² + 2a (-55 m)
a = -7.65 m/s²
Sum of forces in the y direction:
∑F = ma
R − mg = ma
R = m (g + a)
R = (2.0 kg) (9.8 m/s² − 7.65 m/s²)
R = 4.31 N
The distance measured between five successive crests of a wave motion executed by a photon of an electromagnetic radiation is 2.4cm.
1. Determine the wavelength in micrometers
2. what is the frequency of the photon
3. what is the energy possessed by the photon.
Answer:
1. λ = 0.48 cm = 4800 μm
2. υ = 6.25 x 10¹⁰ Hz
3. E = 4.14 x 10⁻²³ J
Explanation:
1.
Since, the wavelength is defined as the distance between two consecutive or successive crests or troughs. Therefore, in this case the wavelength will be equal to:
Wavelength = λ = Distance between 5 successive crests/5
λ = 2.4 cm/5
λ = 0.48 cm = 4800 μm
2.
The frequency of photon can be given as:
υ = c/λ
where,
υ = frequency of photon = ?
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 0.48 cm = 0.0048 m
Therefore,
υ = (3 x 10⁸ m/s)/(0.0048 m)
υ = 6.25 x 10¹⁰ Hz
3.
Now, the energy of photon is given as:
E = hυ
where,
E = Energy = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(6.25 x 10¹⁰ Hz)
E = 4.14 x 10⁻²³ J
An MRI machine needs to detect signals that oscillate at very high frequencies. It does so with an LC circuit containing a 15mH coil. To what value should the capacitance be set to detect a 450 MHz signal?
Answer:
The capacitance is [tex]C = 3.2 9 *10^{-16} \ F[/tex]
Explanation:
From the question we are told that
The induction of the LC circuit is [tex]L = 15 mH = 15 *10^{-3} \ H[/tex]
The frequency is [tex]w = 450 \ MHz = 450 *10^{6} \ Hz[/tex]
The natural frequency is mathematically represented as
[tex]w = \frac{1}{\sqrt{LC} }[/tex]
Where C is the capacitance So
=> [tex]C = \frac{1}{L * w^2}[/tex]
substituting values
[tex]C = \frac{1}{15 *10^{-3} * [450 *10^{6}]^2}[/tex]
[tex]C = 3.2 9 *10^{-16} \ F[/tex]
The value for which the capacitance should be set to detect a 450 MHz signal is [tex]8.34 \times 10^{-24} \;F[/tex]
Given the following data:
Inductance = 15 mH = [tex]15 \times 10^{-3}\;H[/tex]Frequency = 450 MHz = [tex]450 \times 10^6 \;Hz[/tex]To determine the value for which the capacitance should be set to detect a 450 MHz signal:
Mathematically, natural frequency is given by the formula:
[tex]f_o = \frac{1}{2\pi \sqrt{LC} }[/tex]
Where:
L is the inductance.C is the capacitance.Making C the subject of formula, we have:
[tex]C = \frac{1}{(2\pi f_o)^2L} \\\\C = \frac{1}{(2\;\times \;3.142 \times \;450 \;\times\; 10^9)^2 \; \times \;15 \times 10^{-3}}\\\\C = \frac{1}{8 \times 10^{24} \;\times \;15 \times 10^{-3} } \\\\C = \frac{1}{1.2 \times 10^{23}} \\\\C= 8.34 \times 10^{-24} \;F[/tex]
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Two large, flat, horizontally oriented plates are parallel to each other, a distance d apart. Half-way between the two plates the electric field field has a magnitude E. If the separation of the plates is reduced to d/2 what is the magnitude of the electric field half-way between the plates
Answer:
Explanation:
Let the potential difference between the plate is V . Then in the first case
Electric field E between plate
E₁ = V / d
where d is separation between plate
When the plate separation becomes d / 2
Electric field E between plate
E₂ = V / d /2
= 2 V / d =2E₁
Or twice the earlier field
An automotive air conditioner produces a 1-kW cooling effect while consuming 0.75 kW of power. What is the rate at which heat is rejected from this air conditioner
Answer:
The rejected by the air conditioning system is 1.75 kilowatts.
Explanation:
A air conditioning system is a refrigeration cycle, whose receives heat from cold reservoir with the help of power input before releasing it to hot reservoir. The First Law of Thermodynamics describes the model:
[tex]\dot Q_{L} + \dot W - \dot Q_{H} = 0[/tex]
Where:
[tex]\dot Q_{L}[/tex] - Heat rate from cold reservoir, measured in kilowatts.
[tex]\dot Q_{H}[/tex] - Heat rate liberated to the hot reservoir, measured in kilowatts.
[tex]\dot W[/tex] - Power input, measured in kilowatts.
The heat rejected is now cleared:
[tex]\dot Q_{H} = \dot Q_{L} + \dot W[/tex]
If [tex]\dot Q_{L} = 1\,kW[/tex] and [tex]\dot W = 0.75\,kW[/tex], then:
[tex]\dot Q_{H} = 1\,kW + 0.75\,kW[/tex]
[tex]\dot Q_{H} = 1.75\,kW[/tex]
The rejected by the air conditioning system is 1.75 kilowatts.
Consider the following spectrum where two colorful lines (A and B) are positioned on a dark background. The violet end of the spectrum is on the left and the red end of the spectrum is on the right. A B 5. (1 point) What is the name for this type of spectrum? 6. (1 point) Transition A is associated with an electron moving between the n= 1 and n= 3 levels. Transition B is associated with an electron moving between the n= 2 and n= 5 levels. Which transition is associated with a photon of longer wavelength?
Answer:
Explanation:
a )
This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .
b ) The wavelength of a photon is inversely proportional to its energy . Photon due to transition between n = 1 and n = 3 will have higher energy than
that due to transition between n = 2 and n = 5 . So the later photon ( B) will have greater wavelength or photon due to transition between n = 2 and n = 5 will have greater wavelength .
1. A força de atração eletrostática é responsável pela união dos íons, onde cargas elétricas de sinais opostos se atraem. Falando a um nível microscópico, tal atração entre os íons acaba produzindo aglomerados com formas geométricas bem definidas, representados pela figura a seguir: Qual a melhor definição para essa estrutura iônica e que composto é formado a partir dela? * 1 ponto cristais de sal / cloreto de potássio. cristal com face plana / cloreto de cálcio. retículo cristalino / cloreto de sódio. retículo de ligações / clorato de sódio.
Answer:
estrutura cristalina / cloreto de sódio
Explicação:
O cloreto de sódio forma a estrutura cristalina da treliça. Nesta estrutura, um átomo de sódio é cercado por 6 íons de carga oposta. Os íons opostos são negativos porque o sódio perde elétrons, o que o torna ião carregado positivamente chamado cátion, de modo que 6 íons negativos estão presentes ao redor do sódio. A estrutura cristalina também é conhecida como cúbico simples, o que significa que eles são distribuídos em três dimensões com igual distância entre eles e formando um ângulo de 90 graus.
A 269-turn solenoid is 102 cm long and has a radius of 2.3 cm. It carries a current of 3.9 A. What is the magnetic field inside the solenoid near its center?
Answer:
Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T
Explanation:
Given;
number of turns of solenoid, N = 269 turn
length of the solenoid, L = 102 cm = 1.02 m
radius of the solenoid, r = 2.3 cm = 0.023 m
current in the solenoid, I = 3.9 A
Magnitude of the magnetic field inside the solenoid near its centre is calculated as;
[tex]B = \frac{\mu_o NI}{l} \\\\[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
[tex]B = \frac{4\pi*10^{-7} *269*3.9}{1.02} \\\\B = 1.293 *10^{-3} \ T[/tex]
Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T
qual charges, one at rest, the other having a velocity of 104 m/s, are released in a uniform magnetic field. Which charge has the largest force exerted on it by the magnetic field?
Answer:
The correct option is D
Explanation:
From the question we are told that
The speed of the first charge is [tex]v = 0 \ m/s[/tex]
This is because it is at rest
The speed of the second charge is [tex]v = 10^{4} \ m/s[/tex]
Generally the force exerted by a magnetic field on a charge is mathematically represented as
[tex]F_q = q * v * B sin \theta[/tex]
Now looking at this above equation we can see that [tex]F_q[/tex] can only be maximum at [tex]\theta = 90 ^o[/tex] and this only obtained when the direction of the charge (i.e its velocity ) is perpendicular to the direction of the magnetic field
so the correct option for this question is D
What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 480 turns, its secondary 8 turns, and the input voltage is 118 V
Answer:
1.97 V
Explanation:
Applying
N1/N2 = V1/V2................... Equation 1
Where N1 = primary turns, N2 = Secondary turns, V1 = primary/input voltage, V2 = secondary/output voltage.
make V2 the subject of the equation
V2 = V1(N2/N1)............. Equation 2
Given: V1 = 118 V, N1 = 480 turns, V2 = 8 turns.
Substitute into equation 2
V2 = 118(8/480)
V2 = 1.97 V.
What is the relationship between the surface area of a parachute and the amount of air resistance it builds up when it is deployed by a sky diver?
Answer:
An open parachute increases the cross-sectional area of the falling skydiver and thus increases the amount of air resistance which he encounters. Once the parachute is opened, the air resistance overwhelms the downward force of gravity.
Explanation:
The larger a parachute, the greater the force.
Hope it helps you in a little way.
select the example that best describes a renewable resource.
A.after a shuttle launch, you can smell the jet fuel for hours.
B.solar panels generate electricity that keeps the satellites running.
C.tractor trailers are large trucks that run on diesel fuel.
D. we use our barbeque every night; it cooks with propane.
Answer:
B.solar panels generate electricity that keeps the satellites running.
Explanation:
Solar panels are a renewable resource because they take energy from the sun.
Each of the boxes starts at rest and is then pulled for 2.0 m across a level, frictionless floor by a rope with the noted force. Which box has the highest final speed
Answer:
Explanation:
d
To shoot a swimming fish when an intense light beam from a laser gun you should aim
Answer
aim directly at the image
Explanation
the light from the laser beam will also bend when it hits the air water interface , so aim directly at the fish
Balls A and B attract each other gravitationally with a force of magnitude F at distance R. If we triple the mass of ball B and triple the separation of the balls to 3R, what is the magnitude of their attractive force now
Answer:
F₂ = 1/3 FExplanation:
Using the law of gravitation of force to solve this question. The law states that the Force of attraction between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distances between them.
Mathematically, F = GMaMb/R² ... 1
G is the gravitational constant
Ma and Mb are the masses of the balls
R is the distance between the balls
If the mass of ball B is tripled and the magnitude of the separation of the balls is increased to 3R, the force between them will be;
F₂ = GMa(3Mb)/(3R)²
F₂ = 3GMaMb/9R² ... 2
Dividing equation 1 by 2 we will have;
F₂/F = (3GMaMb/9R²)/GMaMb/R²
F₂/F = 3GMaMb/9R² * GMaMb/R²
F₂/F = 3/9
F₂/F = 1/3
F₂ = 1/3 F
This shows that the magnitude of the new attractive force is one-third that of the initial attractive force
g An object with mass 1kg travels at 3 m/s and collides with a stationary object whose mass is 0.5kg. The two objects stick together and continue to move. What is the velocity of the two objects together after collision
Answer:
2
Explanation:
since the second object was in it stationary, it velocity is 0 m/s
Calculate the maximum kinetic energy of electrons ejected from this surface by electromagnetic radiation of wavelength 236 nm.
Answer:
Explanation:
Using E= hc/wavelength
6.63 x10^-34 x3x10^8/ 236nm
19.86*10^-26/236*10^-9
=0.08*10^-35Joules